International Journal of Analysis and Applications
ISSN 2291-8639
Volume 1, Number 2 (2013), 106-112
http://www.etamaths.com

A NOTE ON: MULTI-STEP APPROXIMATION SCHEMES FOR

THE FIXED POINTS OF FINITE FAMILY OF

ASYMPTOTICALLY PSEUDOCONTRACTIVE MAPPINGS

MOGBADEMU, ADESANMI ALAO

Abstract. In this paper, using an analytical technique we obtain a strong

convergence for a modified three- step iterative scheme due to Suantai [6] for
asymptotically pseudocontractive mappings in real Banach spaces. Our result

is an improvement and a correction of Rafiq’s [4] results.

1. Introduction

Let E be an arbitrary real Banach Space and let J : E → 2E
∗

be the normalized
duality mapping defined by

J(x) = {f ∈ E∗ :< x,f >= ‖x‖2 = ‖f‖2},∀x ∈ E

where E∗ denotes the dual space of E and < .,. > denotes the generalized duality
pairing between E and E∗. The single-valued normalized duality mapping is de-
noted by j. Let K be a nonempty closed convex subset of E and T : K → K be a
map.
The mapping T is said to be uniformly L- Lipschitzian if there exists a constant
L > 0 such that

‖Tnx−Tny‖≤ L‖x−y‖
for any x,y ∈ K and ∀n ≥ 1.
The mapping T is said to be asymptotically pseudocontractive if there exists a
sequence (kn) ⊂ [1,∞) with limn→∞kn = 1 and for any x,y ∈ K there exists
j(x−y) ∈ J(x−y) such that

< Tnx−Tny,j(x−y) >≤ kn‖x−y‖2,∀n ≥ 1.

The concept of asymptotically pseudocontractive mappings was introduced by Schu
[5].
Ofoedu [3] used the modified Mann iteration process introduced by Schu [5] ,

xn+1 = (1 −αn)xn + αnTnxn n ≥ 0,
to obtain a strong convergence theorem for uniformly Lipschitzian asymptotically
pseudo-contractive mapping in real Banach space setting. This result itself is a
generalization of many of the previous results (see [3] and the references therein).

2010 Mathematics Subject Classification. 47H10, 46A03.
Key words and phrases. Noor iteration; uniformly Lipschitzian; asymptotically pseudocontrac-

tive ; three-step iterative scheme ; Banach spaces.

c©2013 Authors retain the copyrights of their papers, and all
open access articles are distributed under the terms of the Creative Commons Attribution License.

106



A NOTE 107

Recently, Rafiq [4] employed the iterative scheme introduced by Suantai [6] to es-
tablish a strong convergence result for a modified three-step iterative scheme when
dealing with asymptotically pseudocontractive mappings. In fact, he proved the
following theorem :
Theorem 1.1 ([4]). Let K be a nonempty closed convex subset of E, T : K → K be
the asymptotically pseudocontractive mapping with T(K) bounded and the sequence
kn ⊂ [1,∞), limn→∞kn = 1 such that F(T) = {x ∈ K : Tx = x} 6= φ. Further let
T be uniformly continuous and {an}n≥,{bn}n≥0,{cn}n≥0,{a′n}n≥0,
{b′n}n≥0,{c′n}n≥0,{a′′n}n≥0 be real sequences in [0, 1]; an +bn +cn = 1 = a′n +b′n +c′n
satisfying the following conditions: (i) limn→∞(bn + c

′
n) = 0 = limn→∞ b

′
n =

limn→∞ c
′
n = limn→∞a

′′
n (ii)

∑
n≥0(bn + an) = ∞ .

For arbitrary x0 ∈ K, let {xn}∞n=1 be the iterative sequence defined by

xn+1 = anxn + bnT
nyn + cnT

nzn

(1.1) yn = a
′
nxn + b

′
nT

nzn + c
′
nT

nxn.

zn = (1 −a′′n)xn + a
′′Tnxn

Suppose for any ρ ∈ F(T) there exists a strictly increasing function ψ : [0,∞) →
[0,∞) with ψ(0) = 0 such that

< Tnx−ρ,j(x−ρ) >≤ kn‖x−ρ‖2 −ψ(‖x−ρ‖)
for all x ∈ K. Then {xn}n≥0 converges strongly to a fixed point of T .
We observed some mistakes in the proof of the theorem above. For instance, in
equation (10) of Rafiq [4] the author set; dn = ‖Tnyn −Tnxn+1‖, en = ‖Tnzn −
Tnxn+1‖. He further obtained from equations (12) and (14) that limn→∞‖yn −
xn+1‖ = 0, limn→∞‖zn − xn+1‖ = 0. And, using the uniform continuity of T ,
he concluded that dn = limn→∞‖Tnyn − Tnxn+1‖ = 0, en = limn→∞‖Tnzn −
Tnxn+1‖ = 0. This conclusion is, however not correct.
For example, let Tx = 4x ∀x ∈ R and suppose xn+1 = (1 − 1n) ,yn = zn =
(1 + 1

n
) for all n ≥ 1, obviously dn = limn→∞‖Tnyn − Tnxn+1‖ 6= 0, en =

limn→∞‖Tnzn −Tnxn+1‖ 6= 0. Thus, the result of Rafiq [4] needs to be improve.
In this paper, an improvement and a correction to the main result of Rafiq [4] is
presented.
The following lemmas are needed.
Lemma 1.1 [3, 4] . Let E be real Banach Space and J : E → 2E

∗
be the

normalized duality mapping. Then, for any x,y ∈ E
‖x + y‖2 ≤‖x‖2 + 2 < y,j(x + y) >,∀j(x + y) ∈ J(x + y).

Lemma 1.2 [7]. Let {αn} be a non- negative sequence which satisfies the following
inequality

αn+1 ≤ (1 −λn)αn + σn,
where λn ∈ (0, 1),∀n ∈ N,

∑∞
n=1 λn = ∞ and σn = o(λn). Then limn→∞αn = 0.

2. Main results
Theorem 2.1. Let K be a nonempty closed convex subset of E, T : K → K be
asymptotically pseudocontractive and uniformly Lipschitzian map with Lipschitzian
constant L > 0 and the sequence kn ⊂ [1,∞), limn→∞kn = 1 such that F(T) =



108 MOGBADEMU

{x ∈ K : Tx = x} 6= φ. Let {an}n≥0,{bn}n≥0,{cn}n≥0,{a′n}n≥0,
{b′n}n≥0,{c′n}n≥0,{a′′n}n≥0 be real sequences in [0, 1] satisfying :
(i) an + bn + cn = 1 = a

′
n + b

′
n + c

′
n (ii) limn→∞ bn = limn→∞ b

′
n = limn→∞ cn =

0 = limn→∞ c
′
n (iii)

∑
n≥0(bn + cn) = ∞. For arbitrary x0 ∈ K, let {xn}

∞
n=1 be the

iterative sequence defined by (1.1). Suppose for any ρ ∈ F(T) there exists a strictly
increasing function ψ : [0,∞) → [0,∞) with ψ(0) = 0 such that

< Tnxn −ρ,j(xn −ρ) >≤ kn‖xn −ρ‖2 −ψ(‖xn −ρ‖)

for all x ∈ K. Then {xn}n≥0 converges strongly to a fixed point of T .
Proof: From Lemma 1.2, the equation (1.1) and the definition of the asymptoti-
cally pseudocontractive and uniformly Lipschitzian map, we have

‖xn+1 −ρ‖2 = ‖(1 − bn − cn)(xn −ρ) + bn(Tnyn −ρ) + cn(Tnzn −ρ)‖2
≤ (1 − (bn + cn))2‖xn −ρ‖2

+2〈bn(Tnyn −ρ) + cn(Tnzn −ρ),j(xn+1 −ρ)〉
≤ (1 − (bn + cn))2‖xn −ρ‖2

+2bn〈Tnyn −Tnxn+1,j(xn+1 −ρ)〉
+2bn〈Tnxn+1 −ρ,j(xn+1 −ρ)〉
+2cn〈Tnxn+1 −ρ,j(xn+1 −ρ)〉
+2cn〈Tnzn −Tnxn+1,j(xn+1 −ρ)〉

≤ (1 − (bn + cn))2‖xn −ρ‖2

+2bn(kn‖xn+1 −ρ‖2 −ψ(‖xn −ρ‖))
+2bn‖Tnyn −Tnxn+1‖‖xn+1 −ρ‖
+2cn‖Tnzn −Tnxn+1‖‖xn+1 −ρ‖
+2cn(kn‖xn+1 −ρ‖2 −ψ(‖xn −ρ‖))

≤ (1 − (bn + cn))2‖xn −ρ‖2 + 2kn(bn + cn)‖xn+1 −ρ‖2
−2(bn + cn)ψ(‖xn+1 −ρ‖)
+2bn‖Tnyn −Tnxn+1‖‖xn+1 −ρ‖
+2cn‖Tnzn −Tnxn+1‖‖xn+1 −ρ‖

(2.1)

≤ (1 − (bn + cn))2‖xn −ρ‖2 + 2kn(bn + cn)‖xn+1 −ρ‖2
−2(bn + cn)ψ(‖xn+1 −ρ‖)
+2bn‖Tnyn −Tnxn+1‖‖xn+1 −ρ‖
+2cn‖Tnzn −Tnxn+1‖‖xn+1 −ρ‖

≤ (1 − δn)2‖xn −ρ‖2 + 2knδn‖xn+1 −ρ‖2
−2δnψ(‖xn+1 −ρ‖) + 2bnL‖yn −xn+1‖‖xn+1 −ρ‖
+2cnL‖zn −xn+1‖‖xn+1 −ρ‖,

where 0 ≤ δn = bn + cn < 1.
We note that



A NOTE 109

‖yn −xn+1‖ = ‖xn+1 −yn‖
= ‖(1 − bn − cn)(xn −yn) + bn(Tnyn −yn)

+cn(T
nzn −yn)‖

≤ (1 − bn − cn)‖xn −yn‖ + bn‖Tnyn −ρ + ρ−yn‖
+cn‖Tnzn −ρ + ρ−yn‖

≤ (1 − bn − cn)‖xn −yn‖ + bn(1 + L)‖yn −ρ‖
+cn(L‖zn −ρ‖ + ‖yn −ρ‖)

= (1 − bn − cn)‖xn −yn‖ + bn(1 + L)(‖yn −xn + xn −ρ‖)
+cnL(‖zn −xn + xn −ρ‖) + cn(‖yn −xn + xn −ρ‖)

≤ (1 − bn − cn)‖xn −yn‖
+bn(1 + L)(‖yn −xn‖ + ‖xn −ρ‖)
+cnL(‖zn −xn‖ + ‖xn −ρ‖)
+cn(‖yn −xn‖ + ‖xn −ρ‖)

≤ (1 + bnL)‖xn −yn‖ + δn(1 + L)‖xn −ρ‖
+cnL‖zn −xn‖

(2.2)

= (1 + bnL)‖xn −yn‖ + δn(1 + L)‖xn −ρ‖
+cnL(‖zn −ρ + ρ−xn‖)

≤ (1 + bnL)‖xn −yn‖ + δn(1 + L)‖xn −ρ‖
+cnL(‖zn −ρ‖ + ‖ρ−xn‖)

= (1 + bnL)‖xn −yn‖ + [δn(1 + L) + cnL]‖xn −ρ‖
+cnL‖zn −ρ‖

≤ (1 + bnL)‖xn −yn‖ + [δn(1 + L) + cnL]‖xn −ρ‖
+cnL(1 + a

′′
nL)‖xn −ρ‖

= (1 + bnL)‖xn −yn‖ + [δn(1 + L)
+cnL(2 + a

′′
nL)]‖xn −ρ‖.

Observe that

(2.3)

‖xn −yn‖ = ‖yn −xn‖
= ‖(1 − b′n − c′n)xn + b′nTnzn + c′nTnxn −xn‖
= ‖b′n(Tnzn −xn) + c′n(Tnxn −xn)‖
≤ b′nL‖zn −ρ‖ + (b′n + (1 + L)c′n)‖xn −ρ‖
≤ b′nL(1 + a′′nL)‖xn −ρ‖ + (b′n + (1 + L)c′n)‖xn −ρ‖
= [b′nL(1 + a

′′
nL) + (b

′
n + (1 + L)c

′
n)]‖xn −ρ‖.

Substituting (2.3) into (2.2) then,

(2.4) ‖xn+1 −yn‖ ≤ d1n‖xn −ρ‖



110 MOGBADEMU

where d1n = (1 + bnL)b
′
nL(1 + a

′′
nL) + b

′
n + (1 + L)(c

′
n + δn) + cnL(2 + a

′′
nL).

In a similar way

(2.5)

‖zn −xn+1‖ = ‖xn+1 −zn‖
= ‖(1 − bn − cn)(xn −zn) + bn(Tnyn −zn)

+cn(T
nzn −zn)‖

≤ (1 − bn − cn)‖xn −zn‖ + bnL‖yn −ρ‖ + bn‖zn −ρ‖
+cn(1 + L)‖zn −ρ‖

= (1 −δn)‖xn −zn‖ + bnL‖yn −xn + xn −ρ‖
+bn‖zn −ρ‖ + cn(1 + L)‖zn −ρ‖

≤ (1 − δn)(‖xn −ρ‖ + ‖zn −ρ‖) + bnL‖yn −xn‖
+bnL‖xn −ρ‖ + bn‖zn −ρ‖ + cn(1 + L)‖zn −ρ‖

≤ (1 − δn)(‖xn −ρ‖ + (1 + a′′nL)‖xn −ρ‖)
+bnL[b

′
nL(1 + a

′′
nL) + (b

′
n + (1 + L)c

′
n)]‖xn −ρ‖

+bnL‖xn −ρ‖ + bn(1 + a′′nL)‖xn −ρ‖
+cn(1 + L)(1 + a

′′
nL)‖xn −ρ‖

≤ cn(‖xn −ρ‖ + (1 + a′′nL)‖xn −ρ‖)
+bnL[b

′
nL(1 + a

′′
nL) + (b

′
n + (1 + L)c

′
n)]‖xn −ρ‖

+bnL‖xn −ρ‖ + bn(1 + a′′nL)‖xn −ρ‖
+cn(1 + L)(1 + a

′′
nL)‖xn −ρ‖

= d2n‖xn −ρ‖,

where

(2.6)
d2n = cn(2 + a

′′
nL) + bn(1 + L[b

′
nL(1 + a

′′
nL) + (b

′
n + (1 + L)c

′
n)])

+(1 + a′′nL)(bn + cnL(1 + L)).

Substituting (2.3) and (2.5) into (2.1) we have the equation that follows
(2.7)
‖xn+1 −ρ‖2 = (1 − δn)2‖xn −ρ‖2 + 2knδn‖xn+1 −ρ‖2

−2δnψ(‖xn+1 −ρ‖) + 2bnLd1n‖xn −ρ‖‖xn+1 −ρ‖
+2cnLd

2
n‖xn −ρ‖‖xn+1 −ρ‖

≤ (1 − δn)2‖xn −ρ‖2 + 2knδn‖xn+1 −ρ‖2
−2δnψ(‖xn+1 −ρ‖) + bnLd1n(‖xn −ρ‖2 + ‖xn+1 −ρ‖2)
+cnLd

2
n(‖xn −ρ‖2 + ‖xn+1 −ρ‖2)

≤ (1 − δn)2‖xn −ρ‖2 + 2knδn‖xn+1 −ρ‖2
−2δnψ(‖xn+1 −ρ‖) + δnLd1n(‖xn −ρ‖2 + ‖xn+1 −ρ‖2)
+δnLd

2
n(‖xn −ρ‖2 + ‖xn+1 −ρ‖2) + 2δnLd2n.

Setting,

(2.8)
An = (1 − (2knδn + δnL(d1n + d2n)))
Bn = ((1 −δn)2 + δnL(d1n + d2n))

(2.9)
Cn = (2(1 −kn) −δn − 2L(d1n + d2n))
Dn = 2Ld

2
n.

Suppose we set infn≥N
ψ(‖xn+1−ρ‖)
1+‖xn+1−ρ‖2

= r. Then r = 0. If it is not the case, we

assume that r > 0. Let 0 < r < min{1,r}, then ψ(‖xn+1−ρ‖)
1+‖xn+1−ρ‖2

≥ r, i.e.,

(2.10) ψ(‖xn+1 −ρ‖) ≥ r + r‖xn+1 −ρ‖2 ≥ r‖xn+1 −ρ‖2.



A NOTE 111

Since limn→∞knδn = 0, there exists a natural number N0 such that

1

2
< An < 1,

for all n > N0.
Thus equation (2.7) becomes,

(2.11)
‖xn+1 −p‖2 ≤ BnAn‖xn −p‖

2 − 2δn
ψ(‖xn+1−ρ‖)

An
+ δnDn

An
≤ (1 − δnCn)‖xn −p‖2 − 2δnψ(‖xn+1 −ρ‖) + 2δnDn.

Substituting (2.10) into (2.11), we have

(2.12)
‖xn+1 −p‖2 ≤ 1−δnCn1+2δnr ‖xn −p‖

2 + 2δnDn
1+2δnr

≤ (1 − δn
(Cn+2r)
1+2δnr

)‖xn −p‖2 + 2δnDn1+2δnr
Since limn→∞δn = limn→∞Cn = 0, we choose N1 > N0 such that

(Cn+2r)
1+2δnr

> r,

for all n > N1. It follows from (2.12) that

(2.13) ‖xn+1 −p‖2 ≤ (1 −δnr)‖xn −p‖2 + 2δnDn1+2δnr
for all n > N1. If we set bn = ‖xn−ρ‖, It follows from Lemma 1.2 that, limn→∞ bn =
0, which is a contradiction. Thus, there exists an infinite subsequence such that
limn→∞ bnj0+1 = 0. Next, we prove that limn→∞ bnj0+m = 0 by induction. Let

∀ � ∈ (0, 1), choose nj0 > N such that bnj0+1 < �, Cnj0+1 >
ψ(�)
4
, Dnj0+1 <

ψ(�)
2

.
First, we want to prove bnj0+2 < �. Suppose it is not the case. Then bnj0+2 ≥ �,
this implies ψ(bnj0+2) ≥ ψ(�). Using (2.11) we now obtain the following

(2.14)
b2nj0+2 ≤ b

2
nj0+1

− δn
ψ(�)
4
�2 − 2δnψ(�) + 2δn

ψ(�)
2

≤ b2nj0+1 − δnψ(�)
< �2,

which is a contracdiction. Hence bnj0+2 < � holds and inductively we can show that
bnj0+i < �, ∀i ≥ 1 holds. This implies that limn→∞ bn = 0, i.e., limn→∞‖xn−ρ‖ =
0.

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112 MOGBADEMU

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Department of Mathematics, University of Lagos, Lagos Nigeria